Answer to Question #201679 in Physics for minahil

Question

 A 4 𝑘𝑔 chunk of ice is sliding to the right at 20 𝑚/𝑠 in a frictionless floor completely covered with ice. After a while, this moving chunk of ice collides with another 4 𝑘𝑔 chunk of ice which is at rest and sticks to it. (a) After the collision, how high above the floor will the combined chunks go? (b) Which quantities are conserved in this problem? Explain

Answer 1

(a) Let's first find the final velocity of combination of two chunks after the collision from the law of conservation of momentum:

$$m_1v_1+m_2v_2=(m_1+m_2)v_f,$$ $$v_f=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)},$$ $$v_f=\dfrac{4\ kg\cdot20\ \dfrac{m}{s}+0}{(4\ kg+4\ kg)}=10\ \dfrac{m}{s}.$$

Finally, we can find how high above the floor will the combined chunks go from the law of conservation of energy:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$h=\dfrac{v^2}{2g}=\dfrac{(10\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=5.1\ m.$$

(b) There are two quantities that conserved in this problem: momentum and energy. There is a perfectly inelastic collision, therefore the momentum is conserved. Also, the final kinetic energy of chunks converts into the gravitational potential energy. Hence, energy also conserved.